ta có : x=2,5 = 5/2  ;   y= 3/4

    Thay x= 5/2 ; y= 3/4 vào biểu thức 2x + 2xy - y, ta có

             2 . 5/2 + 2 . 5/2 . 3/4 - 3/4

           = 5 . 15/4 - 3/4

           = 75/4 - 3/4

           = 18

A = 2x + 2xy = 2x.(1 + y)

|x| = 2,5 => x = 2,5 hoặc x = -2,5

+) x = 2,5

=> A = 2 . 2,5 . (1 - 3/4) = 5 . 1/4 = 5/4

+) x = -2,5

=> A = 2 . (-2,5) . (1 - 3/4) = -5 . 1/4 = -5/4

Vậy A = 5/4 hoặc A = -5/4 với |x| = 2,5; y = -3/4. 

A= 2x (2,5 ) + 2x2,5 x ( -3/4)

A= 5/4

A = 2x + 2xy = 2x.(1 + y)

|x| = 2,5 => x = 2,5 hoặc x = -2,5

+) x = 2,5

=> A = 2 . 2,5 . (1 - 3/4) = 5 . 1/4 = 5/4

+) x = -2,5

=> A = 2 . (-2,5) . (1 - 3/4) = -5 . 1/4 = -5/4

Vậy A = 5/4 hoặc A = -5/4 với |x| = 2,5; y = -3/4. 

Giups mik vs

Ta có: \(\left\{{}\begin{matrix}3x-y=3z\\2x+y=7z\end{matrix}\right.\)

\(\Leftrightarrow3x-y+2x+y=10z\)

\(\Leftrightarrow5x=10z\)

hay x=2z

Thay x=2z vào biểu thức 3x-y=3z, ta được:

\(3\cdot2z-y=3z\)

\(\Leftrightarrow6z-y=3z\)

hay y=3z

Thay x=2z và y=3z vào biểu thức \(M=\dfrac{x^2-2xy}{x^2+y^2}\), ta được:

\(M=\dfrac{\left(2z\right)^2-2\cdot2z\cdot3z}{\left(2z\right)^2+\left(3z\right)^2}=\dfrac{4z^2-12z^2}{13z^2}=\dfrac{-8z^2}{13z^2}=\dfrac{-8}{13}\)

Vậy: \(M=\dfrac{-8}{13}\)

\(\left\{{}\begin{matrix}3x-y=3z\\2x+y=7z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5x=10z\\3x-y=3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2z\\3.2z-y=3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2z\\y=3.2z-3z=6z-3z=3z\end{matrix}\right.\)

Có: \(M=\dfrac{x^2-2xy}{x^2+y^2}=\dfrac{\left(2z\right)^2-2.2z.3z}{\left(2z\right)^2+\left(3z\right)^2}=\dfrac{4z^2-12z^2}{4z^2+9z^2}=\dfrac{-8z^2}{13z^2}==-\dfrac{8}{13}\)

\(\left\{{}\begin{matrix}3x-y=3z\\2x+y=7z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5x=10z\\3x-y=3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2z\\3.2z-y=3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2z\\y=3.2z-3z=6z-3z=3z\end{matrix}\right.\)

Có: \(M=\dfrac{x^2-2xy}{x^2+y^2}=\dfrac{\left(2z\right)^2-2.2z.3z}{\left(2z\right)^2+\left(3z\right)^2}=\dfrac{4z^2-12z^2}{4z^2+9z^2}=\dfrac{-8z^2}{13z^2}==-\dfrac{8}{13}\)

chịu

A = (3,1 – 2,5)  – (-2,5 + 3,1) = 3,1 – 2,5 + 2,5 – 3,1 = 0

B  = (5,3 – 2,8) – (4 + 5,3) = 5,3 – 2,8 – 4 – 5,3

A=(3,1-2,5)-(-2,5+3,1)=3,1-2,5+2,5-3,1=0

B=(5,3-2,8)-(4+5,3)=5,3-2,8-4-5,3=-6,8

thiếu đề : \(\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}.\)

Bài 2 :

a, Để \(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\frac{4^2-4}{5}\)

\(\Rightarrow\hept{\begin{cases}2x-2\ne0\\x^2-1\ne0\\2x+2\ne0\end{cases}}\Rightarrow\orbr{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)

b,\(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\frac{4x^2-4}{5}\)

\(B=\left[\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x+1\right)\left(x-1\right)}-\frac{x+3}{2\left(x+1\right)}\right].\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(B=\left[\frac{x^2+2x+1}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+2x-3}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(B=\left[\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(B=\frac{4}{2\left(x-1\right)\left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(B=\frac{8}{5}\)

=> giá trị của B ko phụ thuộc vào biến x

bài 1

=\(^{\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x+1\right)^2}\)

=\(\left(2x+1+2x-1\right)^2\)

=\(\left(4x\right)^2\)

=\(16x^2\)

Tại x=100 thay vào biểu thức trên ta có:

16*100^2=1600000

\(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)=\left[\frac{x+1}{2.\left(x-1\right)}+\frac{3}{x^2-1}-\frac{x+3}{2.\left(x+1\right)}\right]\)

\(\Rightarrow\hept{\begin{cases}x\ne1\\x\ne\pm1\\x\ne-1\end{cases}\Rightarrow x\pm1}\)

Vậy để B xác định => x=+-1

      \(2x^2+y^2+2xy-8x-6y+10=0\)

\(\Rightarrow2.\left(2x^2+y^2+2xy-8x-6y+10\right)=0\)

\(\Rightarrow4x^2+2y^2+4xy-16x-12y+20=0\)

\(\Rightarrow\left(4x^2+y^2+16+4xy-8y-16x\right)+\left(y^2-4y+4\right)=0\)

\(\Rightarrow\left(2x+y-4\right)^2+\left(y-2\right)^2=0\left(1\right)\)

Ta có: \(\hept{\begin{cases}\left(2x+y-4\right)^2\ge0\forall x;y\\\left(y-2\right)^2\ge0\forall y\end{cases}\Rightarrow\left(2x+y-4\right)^2+\left(y-2\right)^2\ge0\forall x;y\left(2\right)}\)

Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}2x+y-4=0\\y-2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+y=4\\y=2\end{cases}\Rightarrow}\hept{\begin{cases}2x+2=4\\y=2\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\y=2\end{cases}}}\)

Chúc bạn học tốt.

A= ( 3,1 -2,5 ) - ( -2,5 + 3,1 )

=3,1 -2,5 + 2,5 - 3,1

=0

Ta có

A = 0,6 - 0,6

A = 0

     đ/s là : 0 nha

\(A=0.6-0.6=0\)

hok tot